# How do you solve the inequality abs(2-3x)>=2/3?

Jul 26, 2017

See a solution process below: $\left(- \infty , \frac{4}{9}\right]$; $\left[\frac{8}{9} , + \infty\right)$

#### Explanation:

The absolute value function takes any negative or positive term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

$- \frac{2}{3} \ge 2 - 3 x \ge \frac{2}{3}$

First, subtract $\textcolor{red}{2}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:

$- \frac{2}{3} - \textcolor{red}{2} \ge 2 - \textcolor{red}{2} - 3 x \ge \frac{2}{3} - \textcolor{red}{2}$

$- \frac{2}{3} - \left(\frac{3}{3} \times \textcolor{red}{2}\right) \ge 0 - 3 x \ge \frac{2}{3} - \left(\frac{3}{3} \times \textcolor{red}{2}\right)$

$- \frac{2}{3} - \frac{6}{3} \ge - 3 x \ge \frac{2}{3} - \frac{6}{3}$

$- \frac{8}{3} \ge - 3 x \ge - \frac{4}{3}$

Now, divide each segment by $\textcolor{b l u e}{- 3}$ to solve for $x$ while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:

$\frac{- \frac{8}{3}}{\textcolor{b l u e}{- 3}} \textcolor{red}{\le} \frac{- 3 x}{\textcolor{b l u e}{- 3}} \textcolor{red}{\le} \frac{- \frac{4}{3}}{\textcolor{b l u e}{- 3}}$

$\frac{\frac{- 8}{3}}{\frac{\textcolor{b l u e}{- 3}}{1}} \textcolor{red}{\le} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}} x}{\cancel{\textcolor{b l u e}{- 3}}} \textcolor{red}{\le} \frac{\frac{- 4}{3}}{\frac{\textcolor{b l u e}{- 3}}{1}}$

$\frac{- 8 \times 1}{3 \times \textcolor{b l u e}{- 3}} \textcolor{red}{\le} x \textcolor{red}{\le} \frac{- 4 \times 1}{3 \times \textcolor{b l u e}{- 3}}$

$\frac{- 8}{- 9} \textcolor{red}{\le} x \textcolor{red}{\le} \frac{- 4}{- 9}$

$\frac{8}{9} \textcolor{red}{\le} x \textcolor{red}{\le} \frac{4}{9}$

Or

$x \le \frac{4}{9}$; $x \ge \frac{8}{9}$

Or, in interval notation:

$\left(- \infty , \frac{4}{9}\right]$; $\left[\frac{8}{9} , + \infty\right)$